Sum of Reciprocals of Squares Alternating in Sign

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Theorem

\(\displaystyle \dfrac {\pi^2} {12}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \dfrac {\left({-1}\right)^{n + 1} } {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac 1 {1^2} - \frac 1 {2^2} + \frac 1 {3^2} - \frac 1 {4^2} + \cdots\)


Proof 1

Let $f \left({x}\right)$ be the real function defined on $\left({0 \,.\,.\, 2 \pi}\right)$ as:

$f \left({x}\right) = \begin{cases} \left({x - \pi}\right)^2 & : 0 < x \le \pi \\ \pi^2 & : \pi < x < 2 \pi \end{cases}$

From Fourier Series: Square of x minus pi, Square of pi, its Fourier series can be expressed as:

$(1): \quad f \left({x}\right) \sim \displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos n x} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin n x}\right)$

We have that:

\(\displaystyle f \left({\pi - 0}\right)\) \(=\) \(\displaystyle \left({\pi - \pi}\right)^2\)
\(\displaystyle \) \(=\) \(\displaystyle 0\)
\(\displaystyle f \left({\pi + 0}\right)\) \(=\) \(\displaystyle \pi^2\)

where $f \left({\pi - 0}\right)$ and $f \left({\pi + 0}\right)$ denote the limit from the left and limit from the right respectively of $f \left({\pi}\right)$.


It is apparent that $f \left({x}\right)$ satisfies the Dirichlet conditions:

$(\mathrm D 1): \quad f$ is bounded on $\left({0 \,.\,.\, 2 \pi}\right)$
$(\mathrm D 2): \quad f$ has a finite number of local maxima and local minima.
$(\mathrm D 3): \quad f$ has $1$ of discontinuity, which is finite.


Hence from Fourier's Theorem:

\(\displaystyle f \left({\pi}\right)\) \(=\) \(\displaystyle \frac {f \left({\pi - 0}\right) + f \left({\pi + 0}\right)} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {0 + \pi^2} 2\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 2\)


Thus setting $x = \pi$ in $(1)$:

\(\displaystyle f \left({\pi}\right)\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + \sum_{n \mathop = 1}^\infty \left({\frac {2 \cos n \pi} {n^2} + \left({\frac {\left({-1}\right)^n \pi} n + \frac {2 \left({\left({-1}\right)^n - 1}\right)} {\pi n^3} }\right) \sin n \pi}\right)\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 2 \sum_{n \mathop = 1}^\infty \frac {\cos n \pi} {n^2}\) Sine of Multiple of Pi
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} 4\) \(=\) \(\displaystyle \frac {\pi^2} 3 + \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^n} {n^2}\) Cosine of Multiple of Pi and simplification
\(\displaystyle \leadsto \ \ \) \(\displaystyle -\frac {\pi^2} {12}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^n} {n^2}\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} {12}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}\) changing sign and subsuming into powers of $-1$

$\blacksquare$


Proof 2

\(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n}\right)^2}\) separating odd positive terms from even negative terms
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac 1 {\left({2 n - 1}\right)^2} - \frac 1 4 \sum_{n \mathop = 1}^\infty \frac 1 {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} 8 - \frac 1 4 \times \frac {\pi^2} 6\) Sum of Reciprocals of Squares of Odd Integers, Basel Problem
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2 \left({3 - 1}\right)} {24}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {\pi^2} {12}\)

$\blacksquare$


Proof 3

Let $f \left({x}\right)$ be the real function defined on $\left({-\pi \,.\,.\, \pi}\right)$ as:

$f \left({x}\right) = \pi^2 - x^2$


By Fourier Series: $\pi^2 - x^2$ over $\left({-\pi \,.\,.\, \pi}\right)$:

$\displaystyle \pi^2 - x^2 \sim \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {\cos n x} {n^2}$

for $x \in \left({-\pi \,.\,.\, \pi}\right)$.


Setting $x = 0$:

\(\displaystyle \pi^2 - 0^2\) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac {\cos 0} {n^2}\)
\(\displaystyle \) \(=\) \(\displaystyle \frac {2 \pi^2} 3 + 4 \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac 1 {n^2}\) Cosine of Zero is One
\(\displaystyle \leadsto \ \ \) \(\displaystyle \frac {\pi^2} {12}\) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \left({-1}\right)^{n - 1} \frac 1 {n^2}\) simplification
\(\displaystyle \) \(=\) \(\displaystyle \sum_{n \mathop = 1}^\infty \frac {\left({-1}\right)^{n + 1} } {n^2}\) rearrangement

$\blacksquare$


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